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3

We have to express the function $(x,y)\mapsto f(x,y)$ in terms of the new variables $u$, $v$. To this end we have to invert the defining equations $$u=bx-ay,\quad v=ax+by$$ and obtain $$x(u,v)={bu+av\over a^2+b^2},\quad y(u,v)={-au+bv\over a^2+b^2}\ .$$ Our function $f$ now appears as $$\tilde f(u,v):=f\bigl(x(u,v),y(u,v)\bigr)\ .$$ Using the chain rule we ...


2

If $G$ is bounded, then for $w\in H^1(\Omega)$, $$ |\langle g(v),w\rangle| \le \int_\Omega |g(v)|\cdot|w| dx \le M \|w\|_{L^1(\Omega)}, $$ and you are done. As you can see, there is a lot of room to relax the assumptions on $g$. In fact, one can allow a certain growth of $g(x)$ for $|x|\to\infty$. Then $g$ is well-defined. Since $g \in W^{2,\infty}$ it is ...


2

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dy}{dt}=1+y$ , letting $y(0)=0$ , we have $y=e^t-1$ $\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0(y+1)$ $\dfrac{du}{dt}=x(1+y)+xu=x_0e^{2t}+x_0e^tu$ , we have ...


2

This is a ridiculously bad multiple-choice question. You are right that (a), (c), and (d) are all correct answers. The only way to have (d) as the unique correct answer is to interpret "correct" as "identical in appearance to what was presented in class". The form $$u=C_1(x+iy)+C_2(x-iy)\tag{1}$$ with complex-valued functions $C_1,C_2$, is also correct. ...


1

Hint: What is the value of $y$ on $\partial U$?


1

Indeed, after some playing with Fresnel functions I've constructed the solution: $$2\psi(x,t)=1+\operatorname S\left(\frac x{\sqrt{2\pi t}}\right)+\operatorname C\left(\frac x{\sqrt{2\pi t}}\right)+i\left[\operatorname S\left(\frac x{\sqrt{2\pi t}}\right)-\operatorname C\left(\frac x{\sqrt{2\pi t}}\right)\right],$$ where $\operatorname C(x)$ and ...


1

$s=x$ and $t=x-y$ hense $s_x=1$ ; $s_y=0$ ; $t_x=1$ ; $t_y=-1$ $$u_x=v_s s_x+v_t t_x=v_s+v_t$$ $$u_{xx}=(v_{ss} s_x+v_{st} t_x)+(v_{ts} s_x+v_{tt} t_x)=v_{ss}+v_{st}+v_{ts}+v_{tt}=v_{ss}+2v_{st}+v_{tt}$$ $$u_{xy}=(v_{ss} s_y+v_{st} t_y)+(v_{ts} s_y+v_{tt} t_y)=-v_{st}-v_{tt}$$ $$u_y=v_s s_y+v_t t_y=-v_t$$ $$u_{yy}=-v_{ts} s_y-v_{tt} t_y=v_{tt}$$ ...


1

for (ii) $f(x) = \int \phi(x-y)\chi(y)dy$. Now if $x \in V$ then the integrand is non-zero if $\phi(x-y)$ is nonzero that is if $|x-y| < \delta/2$. Now the last condition holds if $y \in U$. Hence we can write $f(x) = \int_U \phi(x-y)\chi(y)dy = \int_U \phi(x-y)dy = \int \phi(x-y)dy = 1$. The second last inequality holds since the support of the ...


1

This is not the classical sine series because $$ \left. \sin\left[\frac{(2n-1)\pi}{2L}x\right]\right|_{x=L} = \sin\left((2n-1)\frac{\pi}{2}\right)=\left\{\begin{array}{cc}1, & n \mbox{ is odd}\\-1, & n \mbox{ is even}\end{array}\right. \;. $$ You can definitely tell from this fact alone that this is not the classical sine series. You can ...


1

Your PDE is linear and hence can be solved applying the method of characteristics, which for your particular case reads: $$ \frac{\mathrm{d} x }{x} = \frac{\mathrm{d} y}{1+y} = \frac{\mathrm{d} u}{x(1+y) + xu} = -\frac{\mathrm{d} p}{blabla} = -\frac{\mathrm{d} q }{blabla}, \tag{1} $$ where $blabla$ are things I'm not interested in. Note that from $1$st and ...


1

I spotted the following mistake, which invalidates all steps after it: from $$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$$ you should have gotten $$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$$ Instead, you omitted $C$, then multiplied by $e^{4z}$, then brought $C$ back in, un-multiplied. Why don't you do this in a reasonable way? ...


1

Hint: You showed that if the problem is solvable then $\int_{\Omega}fdx=0$, suppose that $\int_{\Omega}fdx=0$ and show that the problem is solvable you need to proof a existence theorem, or use one if you know some. Edit: try to use Fredholm alternative for every $f ∈ L^2(Ω) $the boundary value problem (∗)$ \triangle u = f \,\,in\,\, Ω,\,\,\, ...


1

Suppose that $u \in C_0^\infty(\Omega)$. Then $$\int_\Omega |D^2 u|^2 \, dx = \int_\Omega \sum_{j,k=1}^n (u_{x_jx_k})^2 \, dx = \sum_{j,k=1}^n \int_\Omega u_{x_jx_k} u_{x_j x_k} \, dx.$$ You can integrate by parts twice to get $$\int_\Omega u_{x_jx_k} u_{x_jx_k} \, dx = - \int_\Omega u_{x_jx_kx_j}u_{x_k} \, dx = \int_\Omega u_{x_j x_j}u_{x_kx_k}\, dx$$ ...


1

Introduce the new unknown functions $$u:=\beta f+\gamma g,\qquad s:=f+g\ .$$ Then the system gets decoupled: $$u_t=\alpha u_{xx}+(\beta-\gamma) u,\qquad s_t=\alpha s_{xx}\ .$$ Maybe there will be difficulties when $\beta=\gamma$; see Juan Ospina's answer.


1

This PDE is equivalent to that $$ \frac{d}{dt}u(at,bt,ct)=0,\quad\text{for all $a,b,c\in\mathbb R$}, $$ and it means that $u:\mathbb R^3\smallsetminus\{0\}\to \mathbb R$, is constant on every straight half-line starting from the origin, and not containing it. If fact, the solution is $$ ...


1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$ $\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$ $\dfrac{dz}{dt}=z$ , letting $z(0)=z_0$ , we have $z=z_0e^t=z_0x$ $\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have ...


1

I will provide a solution using Fourier transform. I guess you are familiar with the following notation. Let's say we have the PDE: $ u_t(x,t)=u_{xx}(x,t)-u_x(x,t)$, with the initial condition $u(x,0)=f(x)$. Firstly, we apply Fourier transform ( with respect to the spatial variable $x$). So, we have: $\begin{array}[t]{l}\hat ...


1

For (b). Since $\lambda_n \to +\infty$, we have $\frac{1}{\lambda_n} \to 0$. Hence, from (A.4) and due to the fact that $$ \frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n} \geq 0, $$ we get $$ 0 \leq \int_{\Omega}\left|\nabla w_n\right|^2 \leq \int_{\Omega}\left|\nabla ...


1

The chain rules are: $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = b \frac{\partial f}{\partial u} + a \frac{\partial f}{\partial v} \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial ...


1

Yes, integration by parts on $\mathbb R^n$ means integrate by parts on $B(0,R)$ and then let $R\to\infty$. Often this is done without writing anything down, when the fact that the boundary term, the integral over $\partial B(0,R)$, tends to zero is taken as "obvious". (The unfortunate fact is that this is often more obvious to the writer than to the ...



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